Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 493 and a standard deviation of 115. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 545.2.
P(X > 545.2) =  
Answer as a number accurate to 4 decimal places.

If 14 of the men are randomly selected, find the probability that their mean score is at least 545.2.
P(x > 545.2) =  
Answer as a number accurate to 4 decimal places.

Assume that any probability less than 5% is sufficient evidence to conclude that the preparation course does help men do better. If the random sample of 14 men does result in a mean score of 545.2, is there strong evidence to support the claim that the course is actually effective?

Answer 1

1)

P(X > 545.2) = P( (X - mean ) / SD > (545.2 - mean) / SD )

= P(Z > (545.2 - 493) / 115)

= P( Z > 0.454)

= 1 - P( Z < 0.454)

= 1 - 0.6751

= 0.3249

P(X > 545.2) = 0.3249

2)

P(X > 545.2) = P( (X - mean ) / (SD / n^0.5) > (545.2 - mean) / (SD / n^0.5) )

= P(Z > (545.2 - 493) / (115 / 14^0.5))

= P( Z > 1.698)

= 1 - P( Z < 1.698)

= 1 - 0.9552

= 0.0447

P(X > 545.2) = 0.0447

3)

Since Probability = 0.0447 < 0.05 ( 5% ) , we can conclude that the preparation course does help men do better and there is strong evidence to support the claim that the course is actually effective.